I have the functional form y= β + ε I'm trying to calculate Var(ε|X) and know that Var(ε|X) = E(εε'|X) as usual. But in calculating E(εε'|X), there's a problem I keep running into.

Here's what I did: E(εε'|X)= E(yy'-yβ'-βy'+ββ'|X)

(then expand out the y's)

= E(ββ'+εβ'+ βε'+ εε' + ....|X)

= E(ββ'|X) + E(εβ'|X) + E(βε'|X)+ E(εε'|X) + E(....|X)

But then, you see that I end up with E(εε'|X) within the solution of E(εε'|X) and that shouldn't happen. That can't be possible, right?

Everything else seems correct, what might be the problem?

  • $\begingroup$ I assume you mean the form y=beta'X+eps.You should not expand the y's as the y's come already from an expansion. But I can't really help you if I don't know where you are heading with the calculation. You want the conditional variance of the epsilon in a certain manner? Otherwise E(epseps'|X) is completely fine as you can replace theoretical moments by empirical moments and use the eps which you can back out of your regression. $\endgroup$
    – J3lackkyy
    Feb 19 at 15:52
  • $\begingroup$ Add: you can replace the epsilon by y-beta'X as you did and then drag out the X's (as they are known) and the beta's (as they are supposevly constant). $\endgroup$
    – J3lackkyy
    Feb 19 at 15:59
  • $\begingroup$ No, just y=beta+epsilon. There's no x in the formula but the professor said that the beta is equal to the population mean, mu. I don't know what the end is supposed to look like since it's just a practice exercise $\endgroup$
    – Dee
    Feb 19 at 16:19
  • $\begingroup$ @J3lackkyy When I leave y, I just get E(yy'|X)-betabeta'. I can't remove the y''s because they depend on X but I also don't know that the expectation should remain $\endgroup$
    – Dee
    Feb 19 at 17:15
  • $\begingroup$ As your model is y=beta+epsilon, I am already confused that you want to calculate the conditional variance with respect to X? Could you give me more context, maybe the full exercise description? $\endgroup$
    – J3lackkyy
    Feb 19 at 17:16

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